Another method of solving inequalities is to express the given inequality with zero on the right side and then determine the sign of the resulting function from either side of the root of the function.
The steps are as follows:
Rewrite the inequality so that there is a zero on the right side.Find all linear factors of the function.To find the critical values, set each linear function to zero and solve for x.Determine the sign of the function in the intervals formed by the critical values.The solution will be those intervals in which the function has the correct signs satisfying the inequality.First, we rearrange the inequality with a zero on the right:
x2 − 2x − 3 > 0
which can be factored to give:
\displaystyle{\left({x}+{1}\right)}{\left({x}-{3}\right)}>{0}(x+1)(x−3)>0
Setting both factors to zero, we get:
\displaystyle{\left({x}+{1}\right)}={0}{\quad\text{and}\quad}{\left({x}-{3}\right)}={0}(x+1)=0and(x−3)=0
\displaystyle{x}=-{1}{\quad\text{and}\quad}{x}={3}x=−1andx=3
Therefore the critical values are
\displaystyle{x}=-{1}{\quad\text{and}\quad}{x}={3}x=−1andx=3.
These critical values divide the number line into 3 intervals:
\displaystyle{x}<-{1}x<−1,\displaystyle-{1}<{x}<{3}−1<x<3, and\displaystyle{x}>{3}x>3.Next, we need to determine the sign (plus or minus) of the function in each of the 3 intervals.
For the first interval, \displaystyle{x}<-{1}x<−1,
The value of \displaystyle{\left({x}+{1}\right)}(x+1) will be negative (substitute a few values of \displaystyle{x}x less than \displaystyle-{1}−1 to check),The value of \displaystyle{\left({x}-{3}\right)}(x−3) will also be negativeSo in the interval \displaystyle{x}<-{1}x<−1, the value of the function x2 − 2x − 3 will be
negative × negative = positive
We continue doing this for the other 2 intervals and summarise the results in this table:
Interval\displaystyle{\left({x}+{1}\right)}(x+1)\displaystyle{\left({x}-{3}\right)}(x−3)sign off(x)\displaystyle{x}<-{1}x<−1−−+\displaystyle-{1}<{x}<{3}−1<x<3+−−\displaystyle{x}>{3}x>3+++We are solving for
\displaystyle{\left({x}+{1}\right)}{\left({x}-{3}\right)}>{0}(x+1)(x−3)>0
The intervals that satisfy this inequality will be those where f(x) has a positive sign.
Hence, the solution is: \displaystyle{x}<-{1}x<−1 or \displaystyle{x}>{3}x>3.
